In triangle ABC, angle BAC=45". D is a point on BC such that AD is perpendicular to BC. If BD = 3cm and DC = 2cm, and the area of the triangle ABC is x cm2. find the value of x.
2007-05-29 15:51:15 · 2 answers · asked by Alfred W 3 in Science & Mathematics ➔ Mathematics
This is a question from a Secondary One. Drawing is not allowed.
2007-05-29 16:00:41 · update #1
It might not be a 45 45 90 triangle. At least the question didn't specify. Please prove it if you can.
2007-05-29 16:04:12 · update #2
I am sorry. 45 should be in degree and not ". (I don't know how to type the degree). Helmut, good try and please try again.
2007-05-29 16:57:33 · update #3
try this
AD will divide the triangle and the angle BAC into two
angle BAD/angle CAD =3/2
ang. BAD =27 and CAD = 18
you can easily find the rest angles
in the tri. BAD the ang. BDA is 90 deg.(perpendicular)
tan BAD =BD/AD
tan 27 =3/ AD
0.51=3/AD
AD=5.9
x = .5 * 5.9 * 5=14.75
2007-06-04 10:47:17 · answer #1 · answered by Anonymous · 0⤊ 0⤋
You give BAC as 45", not 45°.
(45")(1°/3600")(Ï rad./180°) = 0.0002181662 rad.
For angles this small, tan(x) = x
2/AD + 3/AD = 0.0002181662
AD = 5/0.0002181662
A = (5/2)(5/0.0002181662)
A = 57,295.78 cm^2
2007-05-29 16:23:04 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋